Author Topic: Dekoyl's Questions (Read 26138 times) Tweet Share

dekoyl · « **Reply #105 on:** October 09, 2009, 12:03:45 am »

Ah right. I can accept that

Thanks Mao and kamil

dekoyl · « **Reply #106 on:** October 11, 2009, 01:46:55 am »

Hmm..
Roots of $P(z)$ are $\left(z+\frac{\sqrt{3}}{2}+{\frac{1}{2}i\right),\left(z-\frac{\sqrt{3}}{2}-\frac{1}{2}i\right),\left(z-\frac{\sqrt{3}}{2}+\frac{1}{2}i\right),\left(z+\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)$

But when placing it on an Argand diagram, why are the points like so:

Do I have a mental block and it's something very obvious or is it a mistake on their behalf?
Oh, and ignore the shaded part; that's just another part to the Q

Thanks!

TrueTears · « **Reply #107 on:** October 11, 2009, 02:18:24 am »

What is P(z) just out of interest?

But from the looks of it, P(z) must have real coefficients because all the roots are complex conjugates which means it's simply just a reflection in the Re(z) axis.

So as long as you found 2 of the solutions (provided their not conjugates) you can sketch the other 2 since it's just a simple reflection.

dekoyl · « **Reply #108 on:** October 11, 2009, 02:23:30 am »

P(z) was $z^4-z^2+1$

dekoyl · « **Reply #109 on:** October 11, 2009, 02:30:57 am »

Ahhhh. Got it now

Thanks TT.

Off to sleeep

TrueTears · « **Reply #110 on:** October 11, 2009, 02:32:09 am »

Quote from: dekoyl on October 11, 2009, 02:30:57 am

Ahhhh. Got it now Thanks TT.

Off to sleeep

lol soz, ignore my last post (I deleted it), I didn't read their graph carefully, looks like those fools made a mistake, they placed the i on the wrong friggin number!

But previous post still applies to the right answer.

dekoyl · « **Reply #111 on:** October 11, 2009, 02:40:34 am »

Yeah I thought so

Cheers man.

dekoyl · « **Reply #112 on:** October 13, 2009, 12:58:10 am »

Given $\int cos^{-1}(x).dx = xcos^{-1}(x).dx - \sqrt{1-x^2} + c$ , show that the area bounded by the curve and the lines $y=\frac{\pi}{2}, y = \pi$ are given by $\frac{5\pi}{2}-2\left[xcos^{-1}(x)-\sqrt{1-x^2}\right]_{-1}^{0}$

What did I do wrong? I did:
When $y=\frac{\pi}{2}, x = \frac{-1}{2}$

When $y=\pi, x = \frac{-3}{2}$

$A=2\int_{\frac{-3}{2}}^{\frac{-1}{2}}\left(\pi - cos^{-1}\left(x+\frac{1}{2}\right)\right)dx$

$=2\left[\pi x\right]_{\frac{-3}{2}}^{\frac{-1}{2}} - 2\int_{-1}^{0}cos^{-1}(x)dx$

$2\pi - 2\left[xcos^{-1}(x)-\sqrt{1-x^2}\right]_{-1}^{0}$

If I could somehow slot in a $\frac{\pi}{2}$ , then I could get the $\frac{5\pi}{2}$

Thanks!

kamil9876 · « **Reply #113 on:** October 13, 2009, 01:57:44 am »

your missing a rectangle that has area $\frac{\pi}{2}$ . (The one in between $x=\frac{-1}{2}$ and $x=\frac{1}{2}$ ; $y=\frac{\pi}{2}$ and $y=\pi$ )

dekoyl · « **Reply #114 on:** October 13, 2009, 02:00:01 am »

Ah.. I didn't know they wanted that. Because it said "bounded by the curve" and the curve's domain doesn't reach the 'rectangle's area', I thought we could ignore that.

Thanks kamil!

kamil9876 · « **Reply #115 on:** October 13, 2009, 02:11:14 am »

I think the area bounded by curve $C_1$ , $C_2$ ... refers to the area of the enclosed shape whose perimeter is a subset of $C_1 \cup C_2...$ . This applies here as even though the curve is not defined for that domain, the lines $y=\frac{\pi}{2}$ and $y=\pi$ go over that domain and thus make the enclosed shape.

Also, in general, you are right that say $\int_{\frac{-3}{2}}^{\frac{3}{2}} f(x) dx$ does not exist here. But remember that integral and area are not synonymous, the integral is merely used to find the area so the best strategy is to first see which area you want and then think about which integral is equal to it.

dekoyl · « **Reply #116 on:** October 14, 2009, 10:01:52 pm »

Did they do this correctly? Note that $|b| = 2|a|$

Thanks!

Mao · « **Reply #117 on:** October 14, 2009, 10:47:46 pm »

mmm it looks correct.

bloodboy · « **Reply #118 on:** October 15, 2009, 12:04:17 am »

Lol dekoyl you made my day man. Two days ago I was trying to figure the argand diagram question one out, now I know I was right lol. Btw if anyone wants to know it was Neap 07 I think?

dekoyl · « **Reply #119 on:** October 23, 2009, 03:36:46 pm »

The following to equations give the path of spotlights:
$r=10cos(t)i + 5sin(t)j, t\geq0$
$s=5sin(t)i + 10cos(t)j, t\geq0$
The question asks to find the coordinates and time where they intersect.

When finding the time, I did $10cos(t)=5sin(t)$ and $5sin(t)=10cos(t)$ (same equation)
$tan(t)=2 \implies t = 1.11 s$ .

However, why does the answer have 1.11 as one solution and the other solution is $t = 1.11 + \pi$ ? How do I know they want us to find the two solutions in a whole "period"? (I kinda know what they're doing by looking at the unit circle... gah so hard to explain what I'm confused about :S)

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